Two intersecting Cylinders. Please help

[descartes75]

I am stuck on this problem, hope someone can help. It's been a while since my last calc class.

If two cylinders intersect each other perpendicularly with radii "a" and "b" respectively. I want to know the arc length at the intersection . I am having trouble setting up the integrals.
Any help will be most appreciated.

 

[tkhunny]

Find an orientation such that:

\(\displaystyle x^{2} + y^{2} = a^{2}\)

\(\displaystyle x^{2} + z^{2} = b^{2}\)

Subtraction Gives:

\(\displaystyle z^{2} - y^{2} = b^{2} - a^{2}\)

You should at least get an idea if its shape.

 

[galactus]

I have been trying to post for the last couple of hours, but the site would not come up. Anyway, I was thinking a long these lines perhaps:

I would think that the parametrics may be the way t go.

\(\displaystyle \int_{0}^{b}\left[\sqrt{\left(\frac{dx}{dy}\right)^{2}+\left(\frac{dy}{dy}\right)^{2}+\left(\frac{dz}{dy}\right)^{2}}\right]dy\)

If we have two cylinders, say, \(\displaystyle x^{2}+y^{2}=a^{2}, \;\ y^{2}+z^{2}=b^{2}\)

we can use y as a parameter:

\(\displaystyle x=\sqrt{a^{2}-y^{2}}\)

\(\displaystyle y=y\)

\(\displaystyle z=\sqrt{b^{2}-y^{2}}\)

and the y limits of integration would be from 0 to b.

Try that and see what happens. The integral may be a booger, though.

Try some values for the radii a and b and see what you get.

For instance, if we use a=3 and b=2, we get \(\displaystyle \int_{0}^{2}\sqrt{\frac{36-y^{4}}{y^{4}-13y^{2}+36}}dy\)

Which is tough without technology. But with my trusty Voyage 200, I get 3.253...

 

[descartes75]

Thank you.

The above is correct but I am looking for something that can be automated (numerically solved). I am writing a program that would ask the user for the diamters and then spit out the arc length. This integral is a mess to solve with VB

I tried a lot but in vain to solve this in terms of a and b ....

 

[daon]

Thank you.

The above is correct but I am looking for something that can be automated (numerically solved). I am writing a program that would ask the user for the diamters and then spit out the arc length. This integral is a mess to solve with VB

I tried a lot but in vain to solve this in terms of a and b ....

You mean sometimes programming gets hard? :mrgreen:

Suppose b>=a and the cylinder of radius b lies centered on the x axis, the other cylinder of radius a lies centered on the y-axis.

Our equations are:

\(\displaystyle x^2+z^2=a^2\)
\(\displaystyle y^2+z^2=b^2\)

Which as tkhunny pointed out yields: \(\displaystyle y^2-x^2=b^2-a^2 \ge 0\)

If \(\displaystyle b=a\) then they intersect along \(\displaystyle y=\pm|x|\). So that the arclength will be the sum of the circumferences of two equal ellipses: \(\displaystyle \frac{u^2}{2a^2} + \frac{v^2}{a^2}=1\). \(\displaystyle u,v\) I'm using since the ellipses exists in the plane perpendicular to the cartesian plane rotated \(\displaystyle \frac{\pi}{4}\) and \(\displaystyle \frac{3\pi}{4}\) radians. With a little work its not hard to see this, but if you would like a more in-depth explaination I can do so.

Otherwise, The cross-sectional solution is a hyperbola:

\(\displaystyle \frac{y^2}{r^2}-\frac{x^2}{r^2}=1\) where \(\displaystyle r^2=b^2-a^2\).

The tricky part now is finding z in relation to this above equation and the bounds of integration.

I might come back and think some more after a run

 

[descartes75]

That gets me somewhere, at least on the math side. I am thinking Cartesian isn't the right CS for this.